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3 years ago

Function h is a transformation of the parent exponential function. Which statement is true about this function? h(x) = 2^(x-5)

Mathematics

1 answer:

tangare [24]3 years ago

3 0

Answer: As x approaches positive infinity, h(x) approaches positive infinity

Step-by-step explanation:

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Question is on the photo. Thanks!

Elanso [62]

g(x) = 4x + 2

Step-by-step explanation:

f(x) = x + 2

f(4x) = 4x + 2

8 0

2 years ago

Mason walked on Monday, Wensday, and Friday. Use these clues to find how far he walked each day. These distances were six-eights

krok68 [10]

Answer:

Monday he walked $\frac{1}{4}miles$

Wednesday he walked $\frac{6}{8}miles$

Friday he walked $\frac{1}{6}miles$

Step-by-step explanation:

Given Mason walked on Monday, Wednesday and Friday. These distances were six-eights mile, one-fourth mile, and one-sixth mile. He did not walk the farthest on Monday. He walked less on Friday than Monday. we have to find how far he walked each day.

distances are $\frac{6}{8}, \frac{1}{4}, \frac{1}{6}$ that are 0.75, 0.25 and 0.17 respectively.

Now, he didn't walk farthest on Monday and also walked less on Friday than Monday.

∴ Less distance travelled is $\frac{1}{6}$ which is on friday and then $\frac{1}{4}$ on monday.

Rest distance which is $\frac{6}{8}$ on wednesday.

Hence, Monday he walked $\frac{1}{4}miles$

Wednesday he walked $\frac{6}{8}miles$

Friday he walked $\frac{1}{6}miles$

4 0

3 years ago

Read 2 more answers

Brums [2.3K]

Step-by-step explanation:

to my guess the rule of a reflection will be about the x-axis since only the values of y have changed

7 0

2 years ago

Janet set off on a bicycle trip at 7:30 a.m.

UNO [17]

the answer is 10:20.

Hint for the future: be sure to write in the full problem. I only knew the answer because I had to do it myself, but people need more context if they are going to help you.

3 0

3 years ago

19) Given that f(x)x² - 8x+ 15x² - 25find the horizontal and vertical asymptotes using the limits of the function.A) No Vertical

Tems11 [23]

EXPLANATION

Since we have the function:

$f(x)=\frac{x^2-8x+15}{x^2}$

Vertical asymptotes:

$For\:rational\:functions,\:the\:vertical\:asymptotes\:are\:the\:undefined\:points,\:also\:known\:as\:the\:zeros\:of\:the\:denominator,\:of\:the\:simplified\:function.$

Taking the denominator and comparing to zero:

$x+5=0$

The following points are undefined:

$x=-5$

Therefore, the vertical asymptote is at x=-5

Horizontal asymptotes:

$\mathrm{If\:denominator's\:degree\:>\:numerator's\:degree,\:the\:horizontal\:asymptote\:is\:the\:x-axis:}\:y=0.$ $If\:numerator's\:degree\:=\:1\:+\:denominator's\:degree,\:the\:asymptote\:is\:a\:slant\:asymptote\:of\:the\:form:\:y=mx+b.$ $If\:the\:degrees\:are\:equal,\:the\:asymptote\:is:\:y=\frac{numerator's\:leading\:coefficient}{denominator's\:leading\:coefficient}$ $\mathrm{If\:numerator's\:degree\:>\:1\:+\:denominator's\:degree,\:there\:is\:no\:horizontal\:asymptote.}$ $\mathrm{The\:degree\:of\:the\:numerator}=1.\:\mathrm{The\:degree\:of\:the\:denominator}=1$ $\mathrm{The\:degrees\:are\:equal,\:the\:asymptote\:is:}\:y=\frac{\mathrm{numerator's\:leading\:coefficient}}{\mathrm{denominator's\:leading\:coefficient}}$ $\mathrm{Numerator's\:leading\:coefficient}=1,\:\mathrm{Denominator's\:leading\:coefficient}=1$ $y=\frac{1}{1}$ $\mathrm{The\:horizontal\:asymptote\:is:}$ $y=1$

In conclusion:

$\mathrm{Vertical}\text{ asymptotes}:\:x=-5,\:\mathrm{Horizontal}\text{ asymptotes}:\:y=1$

4 0

1 year ago